SET of 8 "C4" Cams and 8 matching wire gates.
Sizes .3 .4 .5 .75 1 2 3 & 4 and 8 anodised "neutrino" - wire gate karabiners.
Chockstone Forum - Accidents & Injuries
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|Danger Area Below Cliff Rescue
Here's one for the physicists/engineers among you.
One of the big risks with a cliff rescue at Arapiles is of people on the ground getting hit by rocks dislodged by the stretcher or rescuers on the cliff. Most rope-rescues at Arapiles involve lowering to the ground and the tendency is for personnel to get close to where the stretcher is going to come down so they can get into things straight away.
One thing we have to do is try to keep people back at a "safe" distance until the stretcher is down.
My question is, how do we work out a "safe" distance.
In my experience from cleaning climbs, people generally underestimate the area affected by a falling rock when it hits the ground, in particular the distance that fragments can shoot out laterally really stuns people (in some cases literally).
My rule of thumb, which has no basis except that it seems better than ignoring the issue, is that people need to be at least as far from the cliff as a rock
can fall. So, if it's a 30 metre high cliff, people should stay 30 metres away from the fall-line.
What does the physics say?
Let's try one example : If a 9Kg rock free-falls 30 metres, and splits into 3 equal fragments when it hits flat ground, how far laterally do the 3 pieces fly?
I'll have a go at a back-of-the-envelope.
The gravitational energy in a 9kg rock falling 30m is mass * gravitational acceleration * height:
mgh = 9 * 10 * 30 = 2700 kJ.
The splitting into three doesn't really make a difference as you'd also split the kinetic energy into three so the distances are the same.
The rock hitting the ground is a very inelastic collision, lots of energy goes into the boom and the crater and breaking apart the rock. Let me pull a number completely out of my arse and say 50% gets converted into kinetic energy.
1/2 mv^2 = 1350
rearrange and get v = sqrt(2*1350/9) = 17 metres per second.
Maximum range you get from launching a projectile from flat ground at 17 meters per second is
v ^ 2 / g = 17^2 / 10 = 29 meters.
Hey, looks like your rule of thumb came out pretty well!
This also ignores the bouncing and rolling after fragments hit the ground.
Being a bit more formal with the maths, it turns out the maximum distance D a rock could bounce upon hitting the bottom of the cliff, is D = 2 * a * h, where h is the height it fell and a is the percentage of energy converted into kinetic energy in the impact. Assuming a is 50% or 0.5, kierenl's rule of thumb is spot on.
On 31/05/2012 kieranl wrote:
>Here's one for the physicists/engineers among you.
>One of the big risks with a cliff rescue at Arapiles is of people on the
>ground getting hit by rocks dislodged by the stretcher or rescuers on the
>cliff. Most rope-rescues at Arapiles involve lowering to the ground and
>the tendency is for personnel to get close to where the stretcher is going
>to come down so they can get into things straight away.
>One thing we have to do is try to keep people back at a "safe" distance
>until the stretcher is down.
>My question is, how do we work out a "safe" distance.
>In my experience from cleaning climbs, people generally underestimate
>the area affected by a falling rock when it hits the ground, in particular
>the distance that fragments can shoot out laterally really stuns people
>(in some cases literally).
>My rule of thumb, which has no basis except that it seems better than
>ignoring the issue, is that people need to be at least as far from the
>cliff as a rock
>can fall. So, if it's a 30 metre high cliff, people should stay 30 metres
>away from the fall-line.
>What does the physics say?
>Let's try one example : If a 9Kg rock free-falls 30 metres, and splits
>into 3 equal fragments when it hits flat ground, how far laterally do the
>3 pieces fly?
Not that I'm that good on the physics of breaking rocks but I think I can safely say that there is no theoretical/analytical answer to that question. With a few assumptions and simplifications you could probably get some fairly useless maximum speeds and such but in a real situation a billion and one slightly different things could happen which would significantly effect the result.
I'll have a go at solving this one scenario which is a bit simpler though as an illustration of what could happen:
10% of the energy gained during the fall was then transferred perfectly into only one of the three fragments, in the form of kinetic energy.
Energy gained during the fall: E = mgh = 9 * 9.8 * 30 ~ 3000J
Energy transfered to one of the three fragmets Efragment = 0.1 * E = 300J
Kinetic energy: Ek = 1/2 * mv2
So velocity of the fragment: vfragment = the square root of (2 * Ek / m) = sqrt (2 * 300 / 3) ~ 14m/s
If this was shot up at an angle of 45 degrees (worst case scenario) and assuming that there is zero air resistance (bad assumption but I definitely couldn't be bothered doing the more realistic calculations) then after a bit more maths of which I couldn't be assed writing up, then it would travel 30m.
Buuuuuuut that calculation relies on some pretty unsafe assumptions - for one, if the rock breaks into smaller fragments then the same amount of energy makes them go faster and further. Then again those small fragments are much more effected by air resistance (imagine dropping a brick and a pebble out of a plane, the brick is less affected by the air resistance).
In reality it comes down to experience and rules of thumb. If your rule of thumb is that people have to be as far away as you are high works then go with that. If people still get occasionally injured, then go further. Beware of creep - if you continually shrink the distance you keep people away by insignificant amounts then eventually people will be close enough to get hurt.
> Not that I'm that good on the physics of breaking rocks but I think I can safely say that there is no theoretical/analytical answer to that question.
With a quick google search it looks like the mining engineers have some models as to how much energy it takes to fracture rock.
Like Olbert said the only real way to be sure is to weigh whole bunch of rocks, drop them from the top of the mount, see what kind of ground they land on, see how far they go, and make some pretty graphs to see if there's a relationship.
Yup, sbm is pretty much on it (except he used kJ instead of J, but it evened out because he reconverted the same way).
If you do all the calcs before substituting any numbers, it very neatly comes out to max horizontal range = 2x starting height. (sbm made a sensible guess of 50% energy dissipation in the impact, which is why it came out as range = starting height).
One disturbing possibility is that upon breaking up on impact, a smaller chunk of the rock can actually gain velocity (as the larger chunk decelerates). You can demonstrate this phenomenon by holding a tennis ball directly on top of a basketball. Now drop both together from head height (get your face out of the way!). The basketball has its bounce deadened quite substantially, while the tennis ball gets launched. If your rock had an ideal bounce and breakup, the no-loss range goes up to about 70m.
One of my co-workers once dropped a steel biner 25m onto one of those cafe shade umbrella's, which acted as the perfect ski jump to launch the thing out of our exclusion zone (it went frigging miles, crossing our large work area, a footpath, and 2 lanes of traffic before it had even bounced!)
I was forgetting about all those people who work in the rigging business. How do you calculate your exclusion zones?
On 31/05/2012 kieranl wrote:
>I was forgetting about all those people who work in the rigging business.
>How do you calculate your exclusion zones?
Ha ha! I always took as much space as I could get away with, but in reality it is never enough to be safe. In canberra the wide footpaths and grassy verges on most of the roads often allowed for 10+m. When I worked in Sydney, it was a joke. The best one I had was in the city, 90m tall building, chunks of decaying concrete, our exclusion zone was about 1.5m wide! In order to be properly safe it would have required closing down a couple of blocks. When I see rope workers on a building, I cross the road.
Edit; I'm not singling out rope workers here, anyone doing work up that high in the city better be pretty good at not dropping stuff (unless they cover the whole building with shadecloth, which they do for major works).
Edit again; None of this stuff is that dangerous, I'd be completely comfortable standing 10m back from a stretcher being lowered as long as I was able to keep my eye on what was happening (I do much more dangerous stuff than that all the time). It's just that when you're having to sign off on it being 'safe' for other people, it kinda sucks because the OH+S morons operate in an unrealistic world. In order for rescues to function, there will be an unavoidable small risk to all of the people involved. I think 20-30m sounds like a good conservative guess which will reduce risk while still allowing your people to be within earshot of the action.
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